Wednesday, July 14, 2010

Problem 472: Triangle, Parallel, Side, Parallelogram, Area, Similarity

Geometry Problem
Click the figure below to see the complete problem 472 about Triangle, Parallel, Side, Parallelogram, Area, Similarity.

Problem 472. Triangle, Parallel, Side, Parallelogram, Area, Similarity
See also:
Complete Problem 472

Level: High School, SAT Prep, College geometry

1 comment:

  1. From N & P draw parallel lines to AB & BC meeting AC in O & Q resply. Hence we've: OH = NP = GQ. As in earlier problems #470 and #471, we can now say that:√(S1/S)=DN/AC=AO/AC,√(S2/S)=PE/AC=QC/AC,√(S3/S)=HG/AC and √(S4/S)=NP/AC= OH/AC = GQ/AC. Thus,(√S1+√S2+√S3+2√S4)/√S =(AO+OH+HG+GQ+QC)/AC = AC/AC = 1 and therefore, √S = √S1 + √S2 + √S3 + 2√S4
    Vihaan

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